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Taking as the definition of 4-force $$F_\mu \equiv \diff{}{\tau} p_\mu$$ then the spatial components of 4-force $\vec F$ look like $$\vec F = \diff{}{\tau} \left[ m \diff{}{\tau} \vec x \right].$$

Now from the definition $\d \tau = \gamma \d t$ we have $$\vec F =  \gamma \diff{}{t} \left[ m \gamma \diff{}{t} \vec x \right] .$$ But $m$ is invariant, hence by the chain rule $$\vec F = \gamma m \left[ \gamma \diffsq{\vec x}{t} + \diff{\vec x}{t} \diff{}{t} ( \gamma ) \right].$$ The derivative of $\gamma$ is $$\diff{}{t} ( 1 - \vec u \cdot \vec u / c^2 )^{-1/2} = -\frac{1}{2} ( 1 - \vec u \cdot \vec u / c^2 )^{-3/2} \diff{}{t} ( - \vec u \cdot \vec u / c^2 ) = \frac{\gamma^3}{c^2} \vec u \cdot \diff{\vec u}{t},$$ so we obtain $$\vec F = \gamma m \left[ \gamma \diffsq{\vec x}{t} + \diff{\vec x}{t}  \frac{\gamma^3}{c^2} \left( \vec u \cdot \diff{\vec u}{t} \right) \right].$$ Finally, identifying $\vec a = \diff{\vec u}{t} = \diffsq{\vec x}{t}$ and $\vec \beta = \vec u / c$: $$\vec F = \gamma m \left[ \gamma \vec a + \gamma^3 \left( \vec \beta \cdot \vec a \right) \dot \vec \beta \right].$$

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